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What is the best way to power 25 red 3watt LED's and 5 blue 3watt LED's with a 100w LED constant current source driver? Answered

Please excuse me as my technical skills are not as good as I would want them to be :D

So, I am trying to build a grow light with this setup:

1. 25 red 3Watt LED's
- DC Forward Voltage (VF) : 2.2-2.8Vdc
- DC Forward Currect (IF) : 700mA

2. 5 blue 3Watt LED's

- DC Forward Voltage (VF) : 3.4-3.8Vdc
- DC Forward Currect (IF) : 700mA

3.100W LED Constant Current Driver

Power: 100W; Input voltage: AC 100~240V 50/60Hz; Output voltage: DC 8~12V; Output current: 8A
(label on the box says: 12V 8A)

What would be the best way to wire the LED's for maximum efficiency? I'm thinking somehow in parallel, but how exactly? Thank you.



Best Answer 3 years ago

If you need to wire them to make use of the ballast shown, like you already have bought it, then that would be pretty difficult to design, and I do not think there is any elegant solution. (Thats the short answer. Read on for a forced solution)

You have 25 red LEDs. I want to spit them up into a few groups of LEDs in series. However, the only factors of 25 are 1, 5, and 25. That means either all LEDs are in series, 5 separate and parallelled groups of LEDs that are in series, and all the EDs in parallel, respectively. Or something altogether different and have some remainders and something more complicated.

The first and last option are not at all possible with that driver, because it either would have horrible efficiency (loads of high value high power resistors dissipating most of the power) or it will not work at all since the driver does not have enough maximum voltage to do it. The option with the 5*5 matrix may be your best option. However, even then there the driver still is capible of more current, but cannot deliver it since the voltage would maxed out at 12V, >3.5A. Especially if you include the balancing resistors.

I have figured this out, not considering the blue LEDs yet, by *ASSUMING* that when LED driver is not loaded down much at all, it goes into CV mode once the maximum 12V output voltage is reached. If that is the case, each LED will have (12V -- unknown Vdrop from current limiting resistor) / (5leds) = <2.4V @ 5LED serieses in parallel = 0.7A * 5 = 3.5A Note: you will need at least a 1 ohm resistor in series with each individual series of LEDs, I do not know exactly how much current the LED draws at <2.4V, probably significantly less than 700mA, and even less so once an equilibrium is reached. (it's really not worth calculating.)


So now we know our supply is going to be outputting 12V, with a remaining 8A-3.5A (which is what all the red LEDs draw)= 4.5A, that should be more than enough to power the remaining blue LEDs with constant voltage mode. We should therefore treat it like a constant voltage source, since it is as long as the current draw is less than 3.5A. CV mode is not ideal for driving LEDs but it can work. It is also not the most elegant solution since 5 is a prime number and cannot be factored, so either all LEDs in parrelell or in series, and like before, either all the LEDs are in series or parrelell, or like 2 LEDs in one series, and 3 in another. It is a messy solution, but the only feasible one. Unless you mix up the red in blue LEDs together in the series, which I guess would not be a big deal, either.


Say 2 of the LEDs are in series across the driver, and another 3 are also, with resistors in series in both cases:


(12V driver)---/\/\/\----l<------l<-----(ground)

(12V driver)---/\/\/\----l<------l<-----l<-----(ground)


If we assume that when 700mA flows through each LED, it will drop 3.6V, then the 2 LED series will drop 7.2V, and the 3 LED series drops 10.8V, then the remainder voltage that has to be gotten rid of for the 2 and 3 LED series is 4.8V and 1.2V respectively. Just do ohms law, knowing that 700mA has to flow, and plug in those voltage drops that have to develop across the resistor, and you should be ready to go. Note, you may also want to use the resistor power formula or watts law to figure out the power dissipated and choose resistors appropriately.

That supply isn't suitable. Its designed for powering one or two very high power LEDs.

You really need a 60W supply with a maximum output of 90V and a constant current of 700mA.

This will underrun them - 600mA, not 700


You can't parallel connect LEDs

So buying 2 of these:


and wiring 15leds in series with the driver, then wiring the other half - 15leds with the driver in series, wiring together the 2 drivers and connecting the to the power plug should do the trick?

That's the simplest plan. Your LEDs will be slightly underpowered, but that will help their longevity.


Please vote for the best answer, to help others. And please post an instructable on the grow light !

How was -Max- answer correct ? You're buying the kind of parts I suggested.

Well, sorry, I wanted to choose your answer, but my question was how to power the LEDs with maximum efficiency with the exact setup given. His detailed, long explanation on the exact question makes it count.


All diodes in parallel and each LED with it's own driver board.

You can't just hook them up with a resistor as the resistor would be huge and only burn heat.

People often think buying high power LED's in bulk will save them money over buying a grow lamp...

can get LED power supplies for up to 72V if you have abosolute
identical LED's that can be put in series to make wiring easier but I
doubt your LED's will it the specs for this.

So you need one of these,


on each LED and all the drivers in parallel on the power supply.

This was not the case, as I would prefer to just buy a ready-to-go grow light, as I already have one. For this one I need a specific, small, compact grow light with my own measurements :)


3 years ago

Sorry that one comment got so large, I was exploring how it would be done if it was the only option. Ignore it all if not. :)


3 years ago

Wiring them in parallel would require many balancing resistors. Each LED I-V curve is a tad bit different, so it is not good practice to wire them in parallel. Also, those resistors will have some voltage drop across them, and that times the current is the power completly lost.

It is best to wire LEDs in series when at all possible. Since all LEDs require the same current, but different voltage drops develop across them when given that current, wiring ALL the LEDs in series, and using a 50-70V ballast (Constant Current LED driver to be specific) that way there are no current limiting resistors used.

If low power factor is not an issue, and electrical isolation from mains on the LEDs is not necessary, using a large, low ESR non polarized capacitor or magnetic ballast to limit the current can actually work, too! However, that is not really the best solution. standard polarized electrolytic caps are out of the question, they will explode. A motor-starting capacitor may work, but I think you'll a need huge high rating MKP/film capacitors. Even then it would be not the safest. A resistor would also work to limit the current, but it is lossy, and the voltage dropped across it times the current is lost power. A proper electronic circuit would be better, but not the cheapest solution.